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Chapter 2: Problem 100

For the following exercises, find vector \(\mathbf{u}\) with a magnitude that is given and satisfies the given conditions. \(\mathbf{v}=\langle 2,4,1\rangle, \quad\|\mathbf{u}\|=15, \quad \mathbf{u}\) and \(\mathbf{v}\) have the same direction

Short Answer

Expert verified
\( \mathbf{u} = \left\langle \pm \frac{10\sqrt{3}}{\sqrt{7}}, \pm \frac{20\sqrt{3}}{\sqrt{7}}, \pm \frac{5\sqrt{3}}{\sqrt{7}} \right\rangle \)

Step by step solution

01

Understand the Direction Condition

Since vector \( \mathbf{u} \) must have the same direction as \( \mathbf{v} \), we know \( \mathbf{u} \) is a scalar multiple of \( \mathbf{v} \). This means \( \mathbf{u} = k \mathbf{v} \) where \( k \) is a real number.
02

Derive the Expression for \( \mathbf{u} \)

If \( \mathbf{u} = k \langle 2, 4, 1 \rangle \), then \( \mathbf{u} \) can be written as \( \langle 2k, 4k, k \rangle \). We will use this expression for the next steps.
03

Use the Magnitude Condition

We know that \( \|\mathbf{u}\| = 15 \), so calculate the magnitude of \( \mathbf{u} \):\[ \|\mathbf{u}\| = \sqrt{(2k)^2 + (4k)^2 + k^2} \]\[ 15 = \sqrt{4k^2 + 16k^2 + k^2} \]\[ 15 = \sqrt{21k^2} \]
04

Solve for \( k \)

Square both sides to solve for \( k \):\[ 225 = 21k^2 \]\[ k^2 = \frac{225}{21} \]\[ k^2 = \frac{75}{7} \]\[ k = \pm \sqrt{\frac{75}{7}} \]
05

Determine the Exact Values of \( k \)

Calculate \( k \):\[ k = \pm \frac{\sqrt{75}}{\sqrt{7}} \]\[ k = \pm \frac{5\sqrt{3}}{\sqrt{7}} \]
06

Construct \( \mathbf{u} \)

Using \( k = \pm \frac{5\sqrt{3}}{\sqrt{7}} \), substitute back into \( \mathbf{u} = k \langle 2, 4, 1 \rangle \) to find:\[ \mathbf{u} = \left\langle \pm \frac{10\sqrt{3}}{\sqrt{7}}, \pm \frac{20\sqrt{3}}{\sqrt{7}}, \pm \frac{5\sqrt{3}}{\sqrt{7}} \right\rangle \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Multiplication
Scalar multiplication involves scaling a vector by multiplying it with a scalar value. In simple terms, each component of the vector is multiplied by the same number. When you apply scalar multiplication to a vector, you essentially stretch or shrink the vector without changing its direction.
In the context of our exercise,
  • we are given that vector \( \mathbf{u} \) is in the same direction as vector \( \mathbf{v} \), meaning \( \mathbf{u} \) is a scaled version of \( \mathbf{v} \).
  • This means \( \mathbf{u} = k \mathbf{v} = k \langle 2, 4, 1 \rangle \), where \( k \) is our scalar.
Scalar multiplication does not affect the direction of \( \mathbf{v} \) unless the scalar has a negative sign, which would reverse the direction. Here, the multiplicative factor \( k \) magnifies or reduces the vector while keeping it pointing in the same direction as \( \mathbf{v} \).
Vector Direction
The direction of a vector is essentially where the vector is 'pointing'. If you imagine a vector as an arrow starting at the origin, pointing in a certain way in space, that is its direction.
In our exercise,
  • vector \( \mathbf{u} \) needs to have the same direction as vector \( \mathbf{v} = \langle 2, 4, 1 \rangle \),
  • so \( \mathbf{u} \) must be some multiple of \( \mathbf{v} \).
This means that for any value of \( k \), positive or negative, as long as \( \mathbf{u} = k \mathbf{v} \), \( \mathbf{u} \) will maintain the direction of \( \mathbf{v} \).In practice, having the same direction implies that each component of \( \mathbf{u} \) is directly proportional to each component of \( \mathbf{v} \) through the scalar \( k \). This characteristic maintains the directionality irrespective of changes in the vector's magnitude.
Magnitude Calculation
Magnitude of a vector is essentially its 'length'. It is computed using the Pythagorean theorem in the context of space dimensions.
To find magnitude,
  • you need to take the square root of the sum of the squares of its components.
  • For vector \( \mathbf{u} = \langle 2k, 4k, k \rangle \), its magnitude is \[ \|\mathbf{u}\| = \sqrt{(2k)^2 + (4k)^2 + k^2}\]
We know from the exercise that \( \|\mathbf{u}\| = 15 \).
After substituting and simplifying,
  • we solve \( 15 = \sqrt{21k^2} \) to find \( k \).
  • Solving the equation gives \( k = \pm \frac{5\sqrt{3}}{\sqrt{7}} \).
This calculation confirms the vector's total length corresponds appropriately to the given magnitude.
Vector Components
Every vector can be broken down into components, which are essentially projections of the vector on the coordinate axes.
In a three-dimensional vector such as \( \mathbf{v} = \langle 2, 4, 1 \rangle \),
  • the first number (2) is the x-component,
  • the second number (4) is the y-component,
  • and the third number (1) is the z-component.
To form \( \mathbf{u} \), which maintains the direction of \( \mathbf{v} \),
  • we multiply each component of \( \mathbf{v} \) by the scalar \( k \),
  • letting \( \mathbf{u} = \langle 2k, 4k, k \rangle \).
Thus, the components of \( \mathbf{u} \) depend on the value of \( k \). Each component demonstrates how much the vector stretches along each axis. These components are crucial when calculating both the direction and magnitude of vector \( \mathbf{u} \).

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Most popular questions from this chapter

For the following exercises, the equation of a plane is given. a. Find normal vector \(\mathbf{n}\) to the plane. Express \(\mathbf{n}\) using standard unit vectors. b. Find the intersections of the plane with the axes of coordinates. c. Sketch the plane. \(x+z=0\)

For the following exercises, lines \(L_{1}\) and \(L_{2}\) are given. Determine whether the lines are equal, parallel but not equal, skew, or intersecting. \(L_{1} : 3 x=y+1=2 z\) and \(L_{2} : x=6+2 t, y=17+6 t, z=9+3 t, \quad t \in \mathbb{R}\)

Consider points \(P(2,1), Q(4,2),\) and \(R(1,2).\) a. Find the area of triangle \(P, Q,\) and \(R.\) b. Determine the distance from point \(R\) to the line passing through \(P\) and \(Q .\)

Consider line \(L\) of symmetric equations \(x-2=-y=\frac{z}{2}\) and point \(A(1,1,1) .\) a. Find parametric equations for a line parallel to L that passes through point A. b. Find symmetric equations of a line skew to L and that passes through point A. c. Find symmetric equations of a line that intersects L and passes through point A.

Consider line \(L\) of parametric equations \(x=t, y=2 t, z=3, \quad t \in \mathbb{R}\) a. Find parametric equations for a line parallel to L that passes through the origin. b. Find parametric equations of a line skew to L that passes through the origin. c. Find symmetric equations of a line that intersects L and passes through the origin.
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